package com.wc.算法提高课.E第五章_数学知识.组合计数.车的放置;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @Author congge
 * @Date 2024/10/7 11:35
 * @description https://www.acwing.com/problem/content/1311/
 */
public class Main {
    /**
     * 思路：<p>
     * 假定一个n * m的矩形, 放置 k 个车, 并且车互相不冲突如何计算? <p>
     * 从 n 行里面选择 k 行 C(n, k), 将这 k 行 排成一列, 第一行有 m 种选择, 第二行有 m - 1种选择... 第 k 行有 m - k + 1种选择, 也就是A(m, k)<p>
     * 那总体就是 C(n, k) * P(m, k)<p>
     * 题目中的图像, 我们将其分为上下两个部分, 一个是 a * b的矩形, 一个是 (a + c) * d的矩形<p>
     * 假定上面 放置 i 个车, 下面就放 k - i个车, 根据上述上面的是 C(b, i) * P(a, i), 下面的就是 C(d, k - i) * (a + c - i, k - i)<p>
     * 因为上面用了 i 列, 所以下面只能使用 a + c - i列了, 得到总体的<p>
     * res = /sum_0_k C(b, i) * P(a, i) * C(d, k - i) * (a + c - i, k - i)<p>
     * C(n, m) = n! / ((n - m)! * m!), P(n, m) = n! / (n - m)! = C(n, m) * m!
     */
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int N = 2010, mod = 100003;
    // fact[i] = i!, infact = fact[i]^-1
    static long[] fact = new long[N], infact = new long[N];
    static int a, b, c, d, k;

    public static void main(String[] args) {
        fact[0] = infact[0] = 1;
        for (int i = 1; i < N; i++) {
            fact[i] = fact[i - 1] * i % mod;
            infact[i] = qmi(fact[i], mod - 2);
        }
        a = sc.nextInt();
        b = sc.nextInt();
        c = sc.nextInt();
        d = sc.nextInt();
        k = sc.nextInt();
        long res = 0;
        for (int i = 0; i <= k; i++) {
            res = (res + C(b, i) * P(a, i) % mod * C(d, k - i) * P(a + c - i, k - i)) % mod;
        }
        out.println(res);
        out.flush();
    }

    static long C(int n, int m) {
        if (n < m) return 0;
        return fact[n] * infact[n - m] % mod * infact[m] % mod;
    }

    static long P(int n, int m) {
        if (n < m) return 0;
        return fact[n] * infact[n - m] % mod;
    }

    static long qmi(long a, int k) {
        long res = 1;
        while (k > 0) {
            if ((k & 1) == 1) res = res * a % mod;
            a = a * a % mod;
            k >>= 1;
        }
        return res;
    }
}

class FastReader {
    StringTokenizer st;
    BufferedReader br;

    FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    String nextLine() {
        String s = "";
        try {
            s = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    // 是否由下一个
    boolean hasNext() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = br.readLine();
                if (line == null)
                    return false;
                st = new StringTokenizer(line);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return true;
    }
}
